HEAT CONDUCTION
OBJECTIVES:
To investigate Fourier’s law of linear conduction
To investigate the temperature profile and heat transfer in radial direction of a
cylinder
To investigate the conduction along a composite bar and evaluate the overall heat
transfer coefficient
To investigate the effect of insulation upon conduction of heat between adjacent
metals
INTRODUCTION:
Heat conduction is the mode of heat transfer in which actual movement of the molecule
doesn’t occur but heat transfers due to the transfer of energy due to vibration of molecules.
Heat conduction takes from higher temperature to lower due to temperature gradient in
the body. It needs a material medium for transfer of heat unlike the radiation.
The conduction of heat is described by the Fourier’s law. According to this law, the heat
flow by conduction in a certain direction is proportional to the area normal to that direction
and to the temperature gradient in that direction.
Mathematically,
dT
Q=−kA
dx
Where, Q = Rate of heat transfer
k = Thermal conductivity of material medium
A = Area of cross section
dT
= Temperature gradient
dx
Negative sign is the indication of transfer of heat towards the body of lower temperature.
Heat is conducted in solid in two ways:
Transport of energy by free electrons and
Lattice vibration.
In good conductors, a large number of free electrons move about in the lattice structure of
the material which transport heat from high temperature region to the lower region. The
portion of energy transported by free electrons is larger than that by lattice vibration. An
increase in temperature causes in both lattice vibration and speed of free electrons, but
increased vibration of lattice disturbs the movement of free electrons which means the
overall conduction s reduced. In insulators and alloys, the transport of energy is mainly due
to lattice vibration and an increase in temperature increases conduction.
Heat is transferred by conduction when adjacent atoms or molecules collide, or as several
electrons move backwards and forwards from atom to atom in a disorganized way so as not
to form a macroscopic electric current, or as phonons collide and scatter. Conduction is the
most significant means of heat transfer within a solid or between solid objects in thermal
contact. Conduction is greater in solids because the network of relatively close fixed spatial
relationships between atoms helps to transfer energy between them by vibration.
Conduction Of Heat Along a simple Bar:
Conduction of heat along a simple conducting bar is shown in the figure below. Let us
consider Fourier’s law of conduction for the case of this bar with lateral surface insulated.
Insulted surface
Q Q
Fig.1. Conduction of heat in a simple insulated
bar
This is an approximation of one-dimensional conduction for a plane wall as shown in Fig.
For steady state condition, it is assumed that the power generated by an electric heater enters
at one end and leaves from the other end and leaves from the other end uniformly. Then the
thermal conductivity of the specimen can be determined as:
Q△ x
k ( T )= w/mk ………(1)
AT
Where, Q = heater power
k (T) = mean value of thermal conductivity between T 1 and T2
T = difference of T1 and T2
APPARATUS REQUIRED
Power switch
Temperature selection switch
Temperature (T)
Insulated brass sample
Brass sample
Heater power socket
Heater power control knob
Fig: Armfield thermal conduction apparatus
Observation table:-
Test No. Wattmete T1 oc T2 oc T3 oc T4 oc T 5 oc T 6 oc T 7 oc T 8 oc T 9 oc
r watts, Q
1 10 46.6 44.4 41.7 36.6 34.3 32.8 22.2 19 18
2 15 47.4 47.1 44.5 37.7 34.5 34.4 22.4 19 17.9
Table 1(for wattmeter reading 10 W) Table 2 (for wattmeter reading 15W)
T(℃ ¿ x(mm) Tx T(℃ ) x(mm) Tx
46.6 10 466 47.4 10 474
44.4 20 888 47.1 20 942
41.7 30 1251 44.5 30 1335
36.6 40 1464 37.7 40 1508
34.3 50 1715 34.5 50 1725
32.8 60 1968 34.4 60 2064
22.2 70 1554 22.4 70 1568
19 80 1520 19 80 1520
18 90 1620 17.9 90 1611
∑T=295.6 ∑x =450 ∑Tx = 12446 ∑T=304.9 ∑x =450 ∑Tx=12747
Calculation:
Specimen material: Brass
Distance between each temperature probes= 10mm
Diameter of specimen (D) = 25mm= 25 *10^-3 m
Length of specimen: 30 mm= 30*10*^-3 m
1 2
πD
Area of cross section (A) = 4
1
π ×(25×10−3 )2
= 4
=4.9087×10-4m2
Calculation of the thermal conductivity of the specimen
−dT
Let a= . The equation of the lines be generalized as
dx
y=ax+ c Where y-axis is temperature axis and x-axis
represents the length of the rod
. i. e . T =ax +T 1 .
Taking summation
∴∑T=𝑎∑𝑥+nX𝑐 …………………………………………………………….………….. (1)
Also, .
xy=a x 2 +cx .
i.e.Tx=a x 2 +c x .
∑ Tx=a ∑ x 2 +c ∑x…………………………........................ (2) [Taking summation]
Thus, using values from table and applying on eqn (1) and eqn (2)
1. For 10W source,
we obtain a=−0.13757
So, dT/dx =−0.13757 K mm-1
= −137 . 57 K m-1
From Fourier’s law of heat conduction,
Q = -K A (dt/dx)
or, K = -Q/(A dt/dx)
= -10/(4.9087×10-4 ×(−137.57))
= 148.08 Wm-1 K-1
2. For 15 W source,
3. 10W,
we obtain a=−0.1 18
So, dT/dx =−0.1 18 K mm-1
= −1 18 K m-1
From Fourier’s law of heat conduction,
Q = -K A (dt/dx)
or, K = -Q/(A dt/dx)
= -10/(4.9087×10-4 ×(−1 18))
= 172.6Wm-1 K-1
Analysis:
In this experiment, we used an Arm field Thermal Conduction Apparatus. In the apparatus
the specimen material used was brass with 9 temperature probes having each temperature
probes at a distance of 10mm. On different power input, we found out the temperature of the
brass rod on different point and calculate the temperature gradient of the rod in order to find
the thermal conductivity of the rod. From the calculation we found two different value of
thermal conductivity at different power input. From the values of temperatures obtained from
the temperature probes connected to the specimen, the value of thermal conductivity of the
material (brass) was calculated.
Both of the values obtained differed from standard values. But we know that the thermal
conductivity of the substance depends only on its material and for brass it should be constant
for a given power input. The fluctuation in the conductivity might be because of the
inaccuracy in experimental procedure, or the insufficient time provided to the apparatus to
gain necessary steady state or the insufficient cooling system. .
The possible cause of the error may be because of It might be because of the malfunctioning
of the temperature probes or due to errors in the procedure.
We also found that heat that flows in certain direction by conduction is directly proportional
to the cross sectional area normal to that direction and the temperature gradient in that
direction.
The Temperature Vs Length graph is shown below.
50
45
t
e 40
m
35
p
r 30
a
25 10 Watt
t 15 Watt
u 20
r
15
e
10
0
0 10 20 30 40 50 60 70 80 90 100 Length
CONCLUSION
From the given experiment, we can conclude that thermal conduction is a mode of heat
transfer in which energy transfer takes place from high temperature region to low
temperature region when a temperature gradient exists in a body. We have also investigated
the Fourier’s law of linear conduction. Similarly, we have studied the nature of the graph
obtained by plotting the temperature of probes versus the distance between the temperature
probes. For the betterment of our result, we should take some precautions during the
experiment.
RECOMMENDATION:
The thermometer should be accurate and must be tightly in contact with the rod so as to
minimize air gap and hence the temperature difference. The rod, 90mm was combined
using three shorter rods of 30mm,during gluing process, there may be air gaps formed that
forms the insulating layer and hence affects the conduction rate. Thus, single rod should be
used as possible. Also, we should not hurry up to take the readings but wait patiently till
steady reading is obtained.
